Question

The total surface area of a hollow metal cylinder open
at both ends of external radius 8 cm and height 10
cm is 3387 cm. Taking r to be inner radius, find the
thickness of the metal in the cylinder.
10 cm​

Answer 1

Answer:

SA of the hollow cylinder = 338π cm²

↠ 2π(8)10 + 2π(r)10 + 2[π(8)² – π(r)²] = 338π

↠ 160 + 20r + 2(64 – r²) = 338

↠ – 2r² + 20r – 50 = 0

↠ r² – 10r + 25 = 0

↠ (r – 5)² = 0

↠ r = 5

∴ Thickness of the metal = 8 – r

= 8 – 5 = 3 cm